A physics book slides off a horizontal table top with a speed of 1.55 m/s . It s

Question

A physics book slides off a horizontal table top with a speed of 1.55 m/s . It strikes the floor after a time of 0.330 s . Ignore air resistance.

1.) Find the height of the table top above the floor.

2.) Find the horizontal distance from the edge of the table to the point where the book strikes the floor.

3.) Find the horizontal component of the book's velocity just before the book reaches the floor.

4.) Find the vertical component of the book's velocity just before the book reaches the floor.

5.) Find the magnitude of the book's velocity just before the book reaches the floor.

6.) Find the direction of the book's velocity just before the book reaches the floor.

Explanation / Answer

along horizantal

x = v0x*t

t = x/v0x

along vertical

initial velocity voy = 0

-y = v0y*t - (1/2)*g*t^2

y = (1/2)*g*t^2

y = (1/2)*g*(x/v0x)^2

1)

y = (1/2)*g*t^2

h = (1/2)*9.8*0.33^2

h = 0.534 m

(2)

x = v0x*t = 1.55*0.33 = 0.512 m

(3)

along horizontal ax = 0

vx = v0x + ax*t = 1.55 m/s

(4)

vy = voy + ay*t

vy = 0 - 9.8*0.33 = -3.234 m/s

(5)

magnitude v = sqrt(vx^2+vy^2) = 3.59 m/s

(6)

direction = tan^-1(vy/vx) = -64.4 degrees

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